Given an array of integers, every element appearstwiceexcept for one. Find that single one.

Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

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Hash Table

Bit Manipulation

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(M) Single Number II

(M) Single Number III

(E) Missing Number

(M) Find the Duplicate Number

(E) Find the Difference

public class Solution {
    public int singleNumber(int[] nums) {
          /**
         * 异或运算符^：两个操作数的位中，相同则结果为0，不同则结果为1
         * 
         * 所以
         * n^n = 0
         * n^0 = n
         */
         int result = 0;
         for(int num : nums){
             result ^= num;
         }
         return result;
    }
}